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Popular >

derivative of tan(ln(ax+b))

Solution

derivative of

tan (ln (a x + b))

Solution

\frac{a sec ^{2} ( ln ( a x + b ) )}{a x + b}

Solution steps

\frac{d}{d x} (tan (ln (a x + b)))

Apply the chain rule:

sec^{2} (ln (a x + b)) \frac{d}{d x} (ln (a x + b))

= sec^{2} (ln (a x + b)) \frac{d}{d x} (ln (a x + b))

\frac{d}{d x} (ln (a x + b)) = \frac{a}{a x + b}

= sec^{2} (ln (a x + b)) \frac{a}{a x + b}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{a sec ^{2} ( ln ( a x + b ) )}{a x + b}

Popular Examples

area y=6-3x^2,y=3x^2,x=-2,x=2 area

y = 6 - 3 x^{2}, y = 3 x^{2}, x = - 2, x = 2

area f(x)=-x^3+8,-8,1 area

f (x) = - x^{3} + 8, - 8, 1

integral of-x^2-x

\int - x^{2} - x d x

integral of (ln(y^3))/(8y)

\int \frac{ln ( y ^{3} )}{8 y} d y

tangent of f(x)=x^2+3x-1,\at x=-1 tangent of

f (x) = x^{2} + 3 x - 1, at x = - 1

Frequently Asked Questions (FAQ)

What is the derivative of tan(ln(ax+b)) ?
The derivative of tan(ln(ax+b)) is (asec^2(ln(ax+b)))/(ax+b)
What is the first derivative of tan(ln(ax+b)) ?
The first derivative of tan(ln(ax+b)) is (asec^2(ln(ax+b)))/(ax+b)